Activity #7 Part I

• Revisit the RailTrail data set from Activity 4.
• Consider two models: a) SLR model to predict ridership by temperature, b) same approach but with added quadratic term.
• Discuss the relative merits of the two models.

For the rest of the week, we won't talk at all about assessing model validity (looking at residual plots). that step is absolutely vital, but we're putting it on hold till next week.

Consider a sample of 15 textbooks. How well can we predict weight by volume?

Consider a sample of 15 textbooks. How well can we predict weight by volume?

summary(m1)

## 
## Call:
## lm(formula = weight ~ volume, data = allbacks)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -190.0 -109.9   38.1  109.7  145.6 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 107.6793    88.3776    1.22     0.24    
## volume        0.7086     0.0975    7.27  6.3e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 124 on 13 degrees of freedom
## Multiple R-squared:  0.803,  Adjusted R-squared:  0.787 
## F-statistic: 52.9 on 1 and 13 DF,  p-value: 6.26e-06

summary(m2)$coef

##             Estimate Std. Error t value  Pr(>|t|)
## (Intercept)  197.963   59.19274   3.344 5.841e-03
## volume         0.718    0.06153  11.669 6.598e-08
## coverpb     -184.047   40.49420  -4.545 6.719e-04

\[ \widehat{weight} = 197.96 + 0.712 volume - 184.047 coverpb \]

• Whichever level is first alphabetically (hb) becomes the reference level.
• coverpb represents the average difference in weight between two books of the same weight but of different covers.
• Again, every row is a t-test that that parameter is zero.

\[ \widehat{weight} = 107.70 + 0.71 volume \]

\[ \widehat{weight}_{hb} = 197.96 + 0.72 volume \\ \widehat{weight}_{pb} = 13.91 + 0.72 volume \]

summary(m1)$r.squared

## [1] 0.8026

summary(m2)$r.squared

## [1] 0.9275

summary(m1)$adj.r.squared

## [1] 0.7875

summary(m2)$adj.r.squared

## [1] 0.9154

We've established that this data is best modeled with two intercepts, but should the two lines have their own slopes as well?

\[ \widehat{weight} = \beta_0 + \beta_1 volume + \beta_2 coverpb + \beta_3 volume * coverpb \]

\[ \widehat{weight}_{hb} = 161.586 + 0.76 volume \\ \widehat{weight}_{pb} = 11.37 + 0.68 volume \]

m3 <- lm(weight ~ volume * cover, data = allbacks)
summary(m3)

## 
## Call:
## lm(formula = weight ~ volume * cover, data = allbacks)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
##  -89.7  -32.1  -21.8   17.9  215.9 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)     161.5865    86.5192    1.87    0.089 .  
## volume            0.7616     0.0972    7.84  7.9e-06 ***
## coverpb        -120.2141   115.6590   -1.04    0.321    
## volume:coverpb   -0.0757     0.1280   -0.59    0.566    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 80.4 on 11 degrees of freedom
## Multiple R-squared:  0.93,   Adjusted R-squared:  0.911 
## F-statistic: 48.5 on 3 and 11 DF,  p-value: 1.24e-06

• Revisit the twins data set from the quiz.
• Is there evidence that the relationship between IQs differs between the social status groups (intercepts or slopes)?